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Posts tagged as “postorder”

LeetCode 145. Binary Tree Postorder Traversal (javascript)

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

Example 1:

Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

Input: root = [1,null,2]
Output: [2,1]

Constraints:

  • The number of the nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Idea:

Use two stacks:

1. Push root to first stack.
2. Loop while first stack is not empty
   2.1 Pop a node from first stack and push it to second stack(res[])
   2.2 Push left and right children of the popped node to first stack
3. Pop all elements from second stack(res.reverse())

Solution:

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var postorderTraversal = function(root) {
    let res = []; // use res as another stack
    // empty tree case:
    if (root === null) return res;
    let stack = [];
    
    // postorder visit: left, right, root
    stack.push(root);
    while (stack.length !== 0) { 
        let cur = stack.pop();
        // treat it as stack, store them in reverse order:
        // i.e root, left, right
        res.push(cur.val); 
        if (cur.left) stack.push(cur.left);
        if (cur.right) stack.push(cur.right);
    }
    
    // we can pop all elements one by one, or just reverse them
    return res.reverse();
};