145. Binary Tree Postorder Traversal - jiejackyzhang/leetcode-note GitHub Wiki

Given a binary tree, return the postorder traversal of its nodes' values.

For example:

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

##Approach 1: Recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        postorderHelper(root, res);
        return res;
    }
    
    private void postorderHelper(TreeNode node, List<Integer> res) {
        if(node == null) return;
        postorderHelper(node.left, res);
        postorderHelper(node.right, res);
        res.add(node.val);
    }
}

##Approach 2: Iterative

postorder为left->right->root，我们可以先按照root->rigth->left的顺序遍历，然后reverse结果即可。 注意到preorder的顺序为root->left->right，因此可以将preorder稍作修改即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            res.add(node.val);
            if(node.left != null) stack.push(node.left);
            if(node.right != null) stack.push(node.right);
        }
        Collections.reverse(res);
        return res;
    }
}
⚠️ **GitHub.com Fallback** ⚠️