145. Binary Tree Postorder Traversal

Link:145

Description

Given a binary tree, return thepostordertraversal 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?

Solution

子问题拆解

1. 需要标记右子树已经visit：lastVisit = node;
2. 不要再traverse左子树，node = null;

过程可视化

打破假设

类比

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    // Solution: Non Recursive
    public List<Integer> postorderTraversal(TreeNode root) {

        List<Integer> postorder = new ArrayList<>();
        if(root == null)
            return postorder;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        TreeNode lastVisit = root;
        while(node != null || !stack.empty()){
            while(node != null){
                stack.push(node);
                node = node.left;
            }
            node = stack.peek();
            if(node.right == null || node.right == lastVisit){
                postorder.add(node.val);
                node = stack.pop();
                lastVisit = node;
                node = null;
            }else{
                node = node.right;
            }
        }
        return postorder;
    }
    /** Soulution 2 recursion
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> postorder = new ArrayList<>();
        traverse(root, postorder);
        return postorder;

    }

    public void traverse(TreeNode node, List<Integer> postorder){
        if(node == null)
            return;
        traverse(node.left, postorder);
        traverse(node.right, postorder);
        postorder.add(node.val);
    }
    */ 
    /** Solution 1 Divide and Conquer
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> postorder = new ArrayList<>();
        if(root == null)
            return postorder;
        List<Integer> left = postorderTraversal(root.left);
        List<Integer> right = postorderTraversal(root.right);
        postorder.addAll(left);
        postorder.addAll(right);
        postorder.add(root.val);
        return postorder;

    }
    */
}/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    // Solution: Non Recursive
    public List<Integer> postorderTraversal(TreeNode root) {

        List<Integer> postorder = new ArrayList<>();
        if(root == null)
            return postorder;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        TreeNode lastVisit = root;
        while(node != null || !stack.empty()){
            while(node != null){
                stack.push(node);
                node = node.left;
            }
            node = stack.peek();
            if(node.right == null || node.right == lastVisit){
                postorder.add(node.val);
                node = stack.pop();
                lastVisit = node;
                node = null;
            }else{
                node = node.right;
            }
        }
        return postorder;
    }
    /** Soulution 2 recursion
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> postorder = new ArrayList<>();
        traverse(root, postorder);
        return postorder;

    }

    public void traverse(TreeNode node, List<Integer> postorder){
        if(node == null)
            return;
        traverse(node.left, postorder);
        traverse(node.right, postorder);
        postorder.add(node.val);
    }
    */ 
    /** Solution 1 Divide and Conquer
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> postorder = new ArrayList<>();
        if(root == null)
            return postorder;
        List<Integer> left = postorderTraversal(root.left);
        List<Integer> right = postorderTraversal(root.right);
        postorder.addAll(left);
        postorder.addAll(right);
        postorder.add(root.val);
        return postorder;

    }
    */
}

Analysis

Reference