本文來自公眾號(hào)程序員小樂(study_tech)

作者：IOExceptioner

編輯：www.jianshu.com/p/0190985635eb

先上二叉樹的數(shù)據(jù)結(jié)構(gòu)：

class?TreeNode{
????int?val;
????//左孩子
????TreeNode?left;
????//右孩子
????TreeNode?right;
}

二叉樹的題目普遍可以用遞歸和迭代的方式來解

1. 求二叉樹的最大深度

int?maxDeath(TreeNode?node){if(node==null){
????????return?0;
????}int?left?=?maxDeath(node.left);int?right?=?maxDeath(node.right);
????return?Math.max(left,right)?+?1;
}

2. 求二叉樹的最小深度

??int?getMinDepth(TreeNode?root){if(root?==?null){return?0;
????????}return?getMin(root);
????}
????int?getMin(TreeNode?root){if(root?==?null){return?Integer.MAX_VALUE;
????????}if(root.left?==?null&&root.right?==?null){return?1;
????????}return?Math.min(getMin(root.left),getMin(root.right))?+?1;
????}

3. 求二叉樹中節(jié)點(diǎn)的個(gè)數(shù)

??int?numOfTreeNode(TreeNode?root){if(root?==?null){
????????????return?0;
????????}int?left?=?numOfTreeNode(root.left);int?right?=?numOfTreeNode(root.right);
????????return?left?+?right?+?1;
????}

4. 求二叉樹中葉子節(jié)點(diǎn)的個(gè)數(shù)

?????int?numsOfNoChildNode(TreeNode?root){if(root?==?null){return?0;
????????}if(root.left==null&&root.right==null){return?1;
????????}return?numsOfNodeTreeNode(root.left)+numsOfNodeTreeNode(root.right);
????}

5. 求二叉樹中第k層節(jié)點(diǎn)的個(gè)數(shù)

??????int?numsOfkLevelTreeNode(TreeNode?root,int?k){if(root?==?null||k<1){return?0;
????????????}if(k==1){return?1;
????????????}int?numsLeft?=?numsOfkLevelTreeNode(root.left,k-1);int?numsRight?=?numsOfkLevelTreeNode(root.right,k-1);return?numsLeft?+?numsRight;
????????}

6. 判斷二叉樹是否是平衡二叉樹

?boolean?isBalanced(TreeNode?node){
????????return?maxDeath2(node)!=-1;
????}int?maxDeath2(TreeNode?node){if(node?==?null){
????????????return?0;
????????}int?left?=?maxDeath2(node.left);int?right?=?maxDeath2(node.right);if(left==-1||right==-1||Math.abs(left-right)>1){
????????????return?-1;
????????}
????????return?Math.max(left,?right)?+?1;
????}

7.判斷二叉樹是否是完全二叉樹

什么是完全二叉樹呢？參見

???boolean?isCompleteTreeNode(TreeNode?root){if(root?==?null){
????????????return?false;
????????}
????????Queue?queue?=?new?LinkedList();
????????queue.add(root);
????????boolean?result?=?true;
????????boolean?hasNoChild?=?false;while(!queue.isEmpty()){
????????????TreeNode?current?=?queue.remove();if(hasNoChild){if(current.left!=null||current.right!=null){
????????????????????result?=?false;
????????????????????break;
????????????????}
????????????}else{if(current.left!=null&&current.right!=null){
????????????????????queue.add(current.left);
????????????????????queue.add(current.right);
????????????????}else?if(current.left!=null&&current.right==null){
????????????????????queue.add(current.left);
????????????????????hasNoChild?=?true;
????????????????}else?if(current.left==null&&current.right!=null){
????????????????????result?=?false;
????????????????????break;
????????????????}else{
????????????????????hasNoChild?=?true;
????????????????}
????????????}
????????}
????????return?result;
????}

8. 兩個(gè)二叉樹是否完全相同

????boolean?isSameTreeNode(TreeNode?t1,TreeNode?t2){if(t1==null&&t2==null){return?true;
????????}else?if(t1==null||t2==null){return?false;
????????}if(t1.val?!=?t2.val){return?false;
????????}
????????boolean?left?=?isSameTreeNode(t1.left,t2.left);
????????boolean?right?=?isSameTreeNode(t1.right,t2.right);return?left&&right;
????}

9. 兩個(gè)二叉樹是否互為鏡像

?????boolean?isMirror(TreeNode?t1,TreeNode?t2){if(t1==null&&t2==null){return?true;
????????}if(t1==null||t2==null){return?false;
????????}if(t1.val?!=?t2.val){return?false;
????????}return?isMirror(t1.left,t2.right)&&isMirror(t1.right,t2.left);
????}

10. 翻轉(zhuǎn)二叉樹or鏡像二叉樹

??????TreeNode?mirrorTreeNode(TreeNode?root){if(root?==?null){
????????????return?null;
????????}
????????TreeNode?left?=?mirrorTreeNode(root.left);
????????TreeNode?right?=?mirrorTreeNode(root.right);
????????root.left?=?right;
????????root.right?=?left;
????????return?root;
????}

11. 求兩個(gè)二叉樹的最低公共祖先節(jié)點(diǎn)

????TreeNode?getLastCommonParent(TreeNode?root,TreeNode?t1,TreeNode?t2){if(findNode(root.left,t1)){if(findNode(root.right,t2)){return?root;
????????????}else{return?getLastCommonParent(root.left,t1,t2);
????????????}
????????}else{if(findNode(root.left,t2)){return?root;
????????????}else{return?getLastCommonParent(root.right,t1,t2)
????????????}
????????}
????}//?查找節(jié)點(diǎn)node是否在當(dāng)前?二叉樹中
????boolean?findNode(TreeNode?root,TreeNode?node){if(root?==?null?||?node?==?null){return?false;
????????}if(root?==?node){return?true;
????????}
????????boolean?found?=?findNode(root.left,node);if(!found){
????????????found?=?findNode(root.right,node);
????????}return?found;
????}

12. 二叉樹的前序遍歷

迭代解法

??ArrayList?preOrder(TreeNode?root){
????????Stack?stack?=?new?Stack();
????????ArrayList?list?=?new?ArrayList();if(root?==?null){return?list;
????????}stack.push(root);while(!stack.empty()){
????????????TreeNode?node?=?stack.pop();list.add(node.val);if(node.right!=null){stack.push(node.right);
????????????}if(node.left?!=?null){stack.push(node.left);
????????????}
????????}return?list;
????}

遞歸解法

??ArrayList?preOrderReverse(TreeNode?root){
????????ArrayList?result?=?new?ArrayList();
????????preOrder2(root,result);return?result;
????}void?preOrder2(TreeNode?root,ArrayList?result){if(root?==?null){return;
????????}
????????result.add(root.val);
????????preOrder2(root.left,result);
????????preOrder2(root.right,result);
????}

13. 二叉樹的中序遍歷

??ArrayList?inOrder(TreeNode?root){
????????ArrayList?list?=?new?ArrayList<();
????????Stack?stack?=?new?Stack();
????????TreeNode?current?=?root;while(current?!=?null||?!stack.empty()){while(current?!=?null){stack.add(current);
????????????????current?=?current.left;
????????????}
????????????current?=?stack.peek();stack.pop();list.add(current.val);
????????????current?=?current.right;
????????}return?list;
????}

14.二叉樹的后序遍歷

??ArrayList?postOrder(TreeNode?root){
????????ArrayList?list?=?new?ArrayList();if(root?==?null){return?list;
????????}list.addAll(postOrder(root.left));list.addAll(postOrder(root.right));list.add(root.val);return?list;
????}

15.前序遍歷和后序遍歷構(gòu)造二叉樹

??TreeNode?buildTreeNode(int[]?preorder,int[]?inorder){if(preorder.length!=inorder.length){return?null;
????????}return?myBuildTree(inorder,0,inorder.length-1,preorder,0,preorder.length-1);
????}TreeNode?myBuildTree(int[]?inorder,int?instart,int?inend,int[]?preorder,int?prestart,int?preend){if(instart>inend){return?null;
????????}
????????TreeNode?root?=?new?TreeNode(preorder[prestart]);int?position?=?findPosition(inorder,instart,inend,preorder[start]);
????????root.left?=?myBuildTree(inorder,instart,position-1,preorder,prestart+1,prestart+position-instart);
????????root.right?=?myBuildTree(inorder,position+1,inend,preorder,position-inend+preend+1,preend);return?root;
????}int?findPosition(int[]?arr,int?start,int?end,int?key){int?i;for(i?=?start;i<=end;i++){if(arr[i]?==?key){return?i;
????????????}
????????}return?-1;
????}

16.在二叉樹中插入節(jié)點(diǎn)

???TreeNode?insertNode(TreeNode?root,TreeNode?node){if(root?==?node){return?node;
????????}
????????TreeNode?tmp?=?new?TreeNode();
????????tmp?=?root;
????????TreeNode?last?=?null;while(tmp!=null){
????????????last?=?tmp;if(tmp.val>node.val){
????????????????tmp?=?tmp.left;
????????????}else{
????????????????tmp?=?tmp.right;
????????????}
????????}if(last!=null){if(last.val>node.val){
????????????????last.left?=?node;
????????????}else{
????????????????last.right?=?node;
????????????}
????????}return?root;
????}

17.輸入一個(gè)二叉樹和一個(gè)整數(shù)，打印出二叉樹中節(jié)點(diǎn)值的和等于輸入整數(shù)所有的路徑

???void?findPath(TreeNode?r,int?i){if(root?==?null){return;
????????}
????????Stack?stack?=?new?Stack();int?currentSum?=?0;
????????findPath(r,?i,?stack,?currentSum);
????}void?findPath(TreeNode?r,int?i,Stack?stack,int?currentSum){
????????currentSum+=r.val;stack.push(r.val);if(r.left==null&&r.right==null){if(currentSum==i){for(int?path:stack){
????????????????????System.out.println(path);
????????????????}
????????????}
????????}if(r.left!=null){
????????????findPath(r.left,?i,?stack,?currentSum);
????????}if(r.right!=null){
????????????findPath(r.right,?i,?stack,?currentSum);
????????}stack.pop();
????}

18.二叉樹的搜索區(qū)間

給定兩個(gè)值 k1 和 k2(k1 < k2)和一個(gè)二叉查找樹的根節(jié)點(diǎn)。找到樹中所有值在 k1 到 k2 范圍內(nèi)的節(jié)點(diǎn)。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找樹的中的節(jié)點(diǎn)值。返回所有升序的節(jié)點(diǎn)值。

??ArrayList?result;ArrayList?searchRange(TreeNode?root,int?k1,int?k2){
????????result?=?new?ArrayList();
????????searchHelper(root,k1,k2);return?result;
????}void?searchHelper(TreeNode?root,int?k1,int?k2){if(root?==?null){return;
????????}if(root.val>k1){
????????????searchHelper(root.left,k1,k2);
????????}if(root.val>=k1&&root.val<=k2){
????????????result.add(root.val);
????????}if(root.val????????????searchHelper(root.right,k1,k2);
????????}
????}

19.二叉樹的層次遍歷

??ArrayList>?levelOrder(TreeNode?root){
????????ArrayList>?result?=?new?ArrayList>();if(root?==?null){return?result;
????????}
????????Queue?queue?=?new?LinkedList();queue.offer(root);while(!queue.isEmpty()){int?size?=?queue.size();
????????????ArrayList<?level?=?new?ArrayList():for(int?i?=?0;i?????????????????TreeNode?node?=?queue.poll();
????????????????level.add(node.val);if(node.left?!=?null){queue.offer(node.left);
????????????????}if(node.right?!=?null){queue.offer(node.right);
????????????????}
????????????}?
????????????result.add(Level);
????????}return?result;
????}

20.二叉樹內(nèi)兩個(gè)節(jié)點(diǎn)的最長距離

二叉樹中兩個(gè)節(jié)點(diǎn)的最長距離可能有三種情況：
1.左子樹的最大深度+右子樹的最大深度為二叉樹的最長距離
2.左子樹中的最長距離即為二叉樹的最長距離
3.右子樹種的最長距離即為二叉樹的最長距離
因此，遞歸求解即可

?private?static?class?Result{??int?maxDistance;??int?maxDepth;??public?Result()?{??
????}??public?Result(int?maxDistance,?int?maxDepth)?{??this.maxDistance?=?maxDistance;??this.maxDepth?=?maxDepth;??
????}??
}??int?getMaxDistance(TreeNode?root){return?getMaxDistanceResult(root).maxDistance;
????}Result?getMaxDistanceResult(TreeNode?root){if(root?==?null){
????????????Result?empty?=?new?Result(0,-1);return?empty;
????????}
????????Result?lmd?=?getMaxDistanceResult(root.left);
????????Result?rmd?=?getMaxDistanceResult(root.right);
????????Result?result?=?new?Result();
????????result.maxDepth?=?Math.max(lmd.maxDepth,rmd.maxDepth)?+?1;
????????result.maxDistance?=?Math.max(lmd.maxDepth?+?rmd.maxDepth,Math.max(lmd.maxDistance,rmd.maxDistance));return?result;
????}

21.不同的二叉樹

給出 n，問由 1…n 為節(jié)點(diǎn)組成的不同的二叉查找樹有多少種？

??int?numTrees(int?n?){int[]?counts?=?new?int[n+2];
????????counts[0]?=?1;
????????counts[1]?=?1;for(int?i?=?2;i<=n;i++){for(int?j?=?0;j????????????????counts[i]?+=?counts[j]?*?counts[i-j-1];
????????????}
????????}return?counts[n];
????}

22.判斷二叉樹是否是合法的二叉查找樹(BST)

一棵BST定義為：
節(jié)點(diǎn)的左子樹中的值要嚴(yán)格小于該節(jié)點(diǎn)的值。
節(jié)點(diǎn)的右子樹中的值要嚴(yán)格大于該節(jié)點(diǎn)的值。
左右子樹也必須是二叉查找樹。
一個(gè)節(jié)點(diǎn)的樹也是二叉查找樹。

?public?int?lastVal?=?Integer.MAX_VALUE;public?boolean?firstNode?=?true;public?boolean?isValidBST(TreeNode?root)?{//?write?your?code?hereif(root==null){return?true;
????????}if(!isValidBST(root.left)){return?false;
????????}if(!firstNode&&lastVal?>=?root.val){return?false;
????????}
????????firstNode?=?false;
????????lastVal?=?root.val;if?(!isValidBST(root.right))?{return?false;
????????}return?true;
????}

祝大家考試順利。

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