100. 相同的树🔖BFS🔖DFS

• https://leetcode-cn.com/problems/same-tree/

BFS

/**
 * 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} p
 * @param {TreeNode} q
 * @return {boolean}
 */
var isSameTree = function(p, q) {
    if (p === null && q === null) return true
    if (p === null || q === null) return false
    
let que_1 = [p]
let que_2 = [q]

while(que_1.length !== 0) {
    let que_1_size = que_1.length
    let que_2_size = que_2.length
    if (que_1_size !== que_2_size) return false
    for (let i = 0; i<que_1_size; i++) {
        let mp1 = que_1.shift()
        let mp2 = que_2.shift()
        if (mp1.val !== mp2.val) return false
        if (mp1.left && mp2.left) {
            que_1.push(mp1.left)
            que_2.push(mp2.left)
        } else if(!mp1.left && !mp2.left){} else {
            return false
        }
        if (mp1.right && mp2.right) {
            que_1.push(mp1.right)
            que_2.push(mp2.right)
        } else if(!mp1.right && !mp2.right){} else {
            return false
        }
    }
}
return true

};

DFS

肝不动了，直接暴力run ok就行

/**
 * 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} p
 * @param {TreeNode} q
 * @return {boolean}
 */
var isSameTree = function(p, q) {
    let result = true
    if (p === null && q === null) return result
    if (p === null || q === null) return !result
    let arr_1 = []
    let arr_2 = []
    let dfs = (data, arr) => {
        if (data) {
            arr.push(data.val)
            if (data.left) {
                dfs(data.left, arr)
            } else {
                arr.push(99999)
            }
            if (data.right) {
                dfs(data.right, arr)
            } else {
                arr.push(99999)
            }
            
    } 
}
dfs(p, arr_1)
dfs(q, arr_2)
if (arr_1.length !== arr_2.length) result = false
arr_1.forEach((data, index) => {
    if (data !== arr_2[index]) {
        result = false
    }
})
return result

};