給定一個(gè)二叉樹(shù)，返回它的 后序?遍歷。

示例:

輸入: [1,null,2,3] ?
? ?1
? ? \
? ? ?2
? ? /
? ?3?

輸出: [3,2,1]

代碼：

迭代：

從根節(jié)點(diǎn)開(kāi)始依次迭代，彈出棧頂元素輸出到輸出列表中，然后依次壓入它的所有孩子節(jié)點(diǎn)，按照從上到下、從左至右的順序依次壓入棧中。

/**

?*?Definition?for?a?binary?tree?node.

?*?public?class?TreeNode?{

?*?????int?val;

?*?????TreeNode?left;

?*?????TreeNode?right;

?*?????TreeNode(int?x)?{?val?=?x;?}

?*?}

?*/

class?Solution?{

????public?List<Integer>?postorderTraversal(TreeNode?root)?{

????????List<Integer>?result?=?new?LinkedList<>();

????????Stack<TreeNode>?stack?=?new?Stack<>();

????????if(root==null){

????????????return?result;

????????}

????????stack.add(root);

????????while(!stack.isEmpty()){

????????????TreeNode?temp?=?stack.pop();

????????????result.add(0,temp.val);

????????????if(temp.left!=null){

????????????????stack.add(temp.left);

????????????}

????????????if(temp.right!=null){

????????????????stack.add(temp.right);

????????????}

????????}

????????return?result;

????}

}

?

遞歸：

/**

?*?Definition?for?a?binary?tree?node.

?*?public?class?TreeNode?{

?*?????int?val;

?*?????TreeNode?left;

?*?????TreeNode?right;

?*?????TreeNode(int?x)?{?val?=?x;?}

?*?}

?*/

class?Solution?{

????List<Integer>?result?=?new?LinkedList<>();

????public?List<Integer>?postorderTraversal(TreeNode?root)?{

????????

????????if(root==null){

????????????return?result;

????????}

????????helper(root);

????????return?result;

????}

????public?void?helper(TreeNode?root){

????????if(root==null){

????????????return;

????????}

????????helper(root.left);

????????helper(root.right);

????????result.add(root.val);

????}

}

總結(jié)

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