二叉树路径总和【一】
shaojinghong opened this issue · comments
景洪 commented
给定一个二叉树和一个目标和,判断该树中是否存在根节点到叶子节点的路径,这条路径上所有节点值相加等于目标和。
说明: 叶子节点是指没有子节点的节点。
原题地址:https://leetcode-cn.com/problems/path-sum/
景洪 commented
可有求所有二叉树路径的算法改进得到:
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {boolean}
*/
var hasPathSum = function(root, sum) {
if (!root) return [];
var stringBuffer = [];
var result = [];
function sumArr(arr){
return arr.reduce(function(prev,cur){
return prev + cur;
}, 0);
}
function visit(root) {
if (!root) return
stringBuffer.push(root.val);
visit(root.left);
visit(root.right);
if (!root.left && !root.right) {
result.push(sumArr(stringBuffer) === sum);
}
//这个是关键步骤
stringBuffer.length --;
}
visit(root);
return result.indexOf(true) !== -1;
}
景洪 commented
左右子树递归法:
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {boolean}
*/
var hasPathSum = function(root, sum) {
if (!root) return false;
if (!root.left && !root.right) {
return root.val - sum == 0;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
};