112. Path Sum

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

ex1

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Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true

ex2

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Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false

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Example 3:

Input: root = [1,2], targetSum = 0
Output: false

Constraints:

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        List<TreeNode> q = new ArrayList<>();
        q.add(root.left);
        q.add(root.right);
        
        while (q.size() > 1) {
            TreeNode left = q.remove(0);
            TreeNode right = q.remove(0);
            
            if (left == null && left == right) {
                continue;
            }
            
            if (left == null || right == null) {
                return false;
            }
            
            if (left.val != right.val) {
                return false;
            }
            
            q.add(left.left);
            q.add(right.right);
            
            q.add(left.right);
            q.add(right.left);
        }
        
        return true;
    }
    
    public boolean isSymmetricRecursive(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        return isSymmetric(root.left, root.right);
    }
    
    boolean isSymmetric(TreeNode left, TreeNode right) {
        
        if (left == null && right == null) {
            return true;
        }
        
        if (left == null || right == null) {
            return false;
        }
        
        if (left.val != right.val) {
            return false;
        }
        
        return isSymmetric(left.left, right.right) 
            && isSymmetric(right.left, left.right);
    }
}