112-Path Sum

Description

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

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Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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問題描述

給定一個二叉樹和sum，判斷二叉樹是否有一條從根節點到葉子節點的路徑， 使得路徑上的所有節點的值之和等於sum

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問題分析

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解法1

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null)    return false;
        if(root.left == null && root.right == null) return sum == root.val;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

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解法2

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        Stack<TreeNode> stack = new Stack();        
        stack.push(root);

        while(!stack.isEmpty() && root != null){
            TreeNode cur = stack.pop() ;    
            if(cur.left == null && cur.right == null){              
                if(cur.val == sum) return true;
            }
            if(cur.right != null){
                cur.right.val = cur.val + cur.right.val ;
                stack.push(cur.right) ;
            }
            if(cur.left != null){
                cur.left.val = cur.val + cur.left.val;
                stack.push(cur.left);
            }
        }       

        return false ;
     }
}