20220504

路径总和(从根节点出发)

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

/**
 * 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)
 * }
 */
 // 解题思路：
路径： 根节点-> 叶子节点
其中叶子节点没有子节点
var pathSum = function(root, targetSum) {
    let ret = []
    const isValid = (node,sum) => {
        if (sum!=targetSum) return false
        if (node.left || node.right) return false 
        return true
    }
    const dfs = (node,sum,path=[]) => {
      if(!node) return
      let _sum = sum + node.val
      if(isValid(node,_sum)) {
        path.push(node.val)
        ret.push(path)
        return 
      }
      if(node.left) dfs(node.left, _sum, [...path, node.val])
      if(node.right) dfs(node.right, _sum, [...path, node.val])
    }
    dfs(root,0)
    return ret
};

路径总和(从任意节点出发)

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3

/**
 * 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)
 * }
 */
// 解题思路：
// 1.遍历树，以每个节点为起始点；
// 2.因为存在后面子节点值之和等于0；故isVaild(_sum)满足条件后，需要继续遍历；
var pathSum = function(root, targetSum) {
    let ret = 0;
    const isVaild = (sum) => {
        return sum == targetSum
    }
    const dfs = (node, sum) => {
        if(!node) return
        let _sum = node.val + sum
        if(isVaild(_sum)) ret++
        dfs(node.left, _sum)
        dfs(node.right, _sum)
    }
    // 遍历树，以每个节点为起始点
    const dfsTree = (node) => {
        if(!node) return
        dfs(node, 0)
        dfsTree(node.left, 0)
        dfsTree(node.right, 0)
    }
    dfsTree(root)
    return ret
};