129. 求根节点到叶节点数字之和🔖DFS

129. 求根节点到叶节点数字之和🔖DFS

https://leetcode-cn.com/problems/sum-root-to-leaf-numbers/dfs/** * Definition for a binary tree node. * function TreeNode(val, left, right) { * ...

2022年5月1日
215字
9 阅读

dfs

/**
 * 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 sumNumbers = function(root) {
    if (!root) return root
    let arr = []
    let result = 0
    let dfs = (data, presum) => {
        if (data) {
            presum = data.val + presum * 10
            if (!data.left && !data.right)
                arr.push(presum)
            else {
                dfs(data.left, presum)
                dfs(data.right, presum)
            }
        }
    }
    dfs(root, 0)
    result = arr.reduce((a, b) => {return a + b})
    return result
};

优化一下得到:

/**
 * 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 sumNumbers = function(root) {
    let dfs = (data, presum) => {
        if (data === null) return 0
        presum = data.val + presum * 10
        if (!data.left && !data.right) {
            return presum
        }
        return dfs(data.left, presum) + dfs(data.right, presum)
    }
    return dfs(root, 0)
};

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